Friday, July 17, 2020

Symmetry, regularity and qualitative probability

Let ⪅ be a qualitative probability comparison for some collection F of subsets of a space Ω. Say that A ≈ B iff A ⪅ B and B ⪅ A, and that A < B provided that A ⪅ B but not B ⪅ A. Minimally suppose that ⪅ is a partial preorder (i.e., transitive and reflexive). Say it’s total provided that for all A and B either A ⪅ B or B ⪅ A. Suppose that G is a group of symmetries acting on Ω, and that F is G-invariant in the sense that gA ∈ F for all g ∈ G. Then we can define:

  1. ⪅ is strongly G-invariant provided that for all A in G and all g in G we have A ≈ gA, and

  2. ⪅ is weakly G-invariant provided that for all A and B in G and all g in G we have A ⪅ B iff gA ⪅ gB.

There is some reason to be suspicious of strong G-invariance. For in some interesting cases, say where Ω is a circle that G is the set of all rotations, there will be cases where gA is a proper subset of A, and by regularity we would expect to have gA < A rather than gA ≈ A. But weak G-invariance seems harder to question.

Say that g ∈ G is of order n provided that gn = e, where e is identity. However, we also have:

Lemma 1. If ⪅ is total, g is of order 2, and A ⪅ B implies gA ⪅ gB for all A and B, then A ≈ gA for all A.

Proof: Since ⪅ is a total order, either A ⪅ gA or gA ⪅ A. Suppose A ⪅ gA. Then gA ⪅ g2A. But g2A = A. Hence gA ≈ A. Similarly, if gA ⪅ A, then g2A ≈ gA. But g2A = A, so gA ≈ A.

So, we have:

Proposition 1. If ⪅ is total and G is a group generated by elements of order 2, then weak G-invariance entails strong G-invariance.

Say that ⪅ is strongly regular provided that if A is a proper subset of B, then A < B. Weak regularity would say that if B is non-empty then ∅ < B. Weak regularity together with an appropriate additivity condition will imply strong regularity (details left to the reader).

Proposition 2. If G is generated by elements of order 2, and ⪅ is total and weakly G-invariant, then if there exists a g ∈ G and A ∈ F such that gA is a proper subset of A, then G is not strongly regular.

Proof: Strong regularity would require that gA < A, but that would contradict strong G-invariance which we have by Proposition 1.

Corollary 1. If F is a collection of subsets of the unit circle containing all countable sets and invariant under all reflections, and ⪅ is a total qualitative probability comparison weakly invariant under all reflections, then ⪅ is not strongly regular.

Proof: The group generated by all reflections includes all rotations. But there is a subset A of the circle and a rotation g such that gA is a proper subset of A. For instance, let A be the set of points at angles r, 2r, 3r, ... in degrees, where r is irrational, and let g be rotation by r. Then rA is the set of points at angles 2r, 3r, 4r, ... in degrees.

Now, imagine an infinite line on which there are infinitely many evenly spaced people, stretching out in both directions, each of whom flips a fair coin. Let Ω be the probability space describing these flips. Let Hn be the event that all the flips starting with person number n (i.e., n, n + 1, n + 2, ...) land heads. Suppose that F contains all the Hn and is invariant under all reflections of the situation (where we reflect the setup either about the point at which some person stands or at a point half-way between two neighboring people).

Corollary 2. If ⪅ is a total qualitative probability comparison weakly invariant under all reflections, then ⪅ is not strongly regular.

Proof: Let G be the group generated by the reflections. This group contains all translations. A non-trivial translation of Hn will either be a proper subset or a proper superset of Hn, depending on the direction of Hn. So by Proposition 2 we cannot have regularity.

Bibliographic note: Lemma 1 and Corollary 2 are analogous to Lemma 2 and Theorem 4 of this paper.

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